Q15 of 95 Page 5

Find the length of altitude AD of an isosceles Δ ABC in which AB = AC = 2a units and BC = a units.


Given: ABC is an isosceles triangle


∴ AB = AC = 2a and BC = a


and AD is the altitude on BC. Therefore, BC = 2BD


Now, In ∆ADB, using Pythagoras theorem, we have


(Perpendicular)2 + (Base)2 = (Hypotenuse)2


⇒ (AD)2 + (BD)2 = (AB)2








[taking positive square root]


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